Question

suppose w(t) denotes the amount of a radioactive material left after time t. assume that w(0)=9 and w(1)=3. complete parts (a) through (c).
(a) find the differential equation that describes this situation.
the differential equation is \\( \frac{dw}{dt}=(\square)w(t) \\)

Explanation:

Step1: Recall the general form of exponential decay

The general form of exponential decay is \(W(t)=W(0)e^{kt}\), where \(W(0)\) is the initial amount and \(k\) is the decay constant. Given \(W(0) = 9\), so \(W(t)=9e^{kt}\).

Step2: Use the given condition \(W(1) = 3\) to find \(k\)

Substitute \(t = 1\) and \(W(1)=3\) into \(W(t)=9e^{kt}\).
We get \(3 = 9e^{k\times1}\), then \(e^{k}=\frac{3}{9}=\frac{1}{3}\).
Take the natural logarithm of both sides: \(k=\ln(\frac{1}{3})=-\ln 3\).

Step3: Recall the differential equation for exponential decay

The differential equation for \(W(t)=W(0)e^{kt}\) is \(\frac{dW}{dt}=kW(t)\). Since \(k =-\ln 3\), the differential equation is \(\frac{dW}{dt}=(-\ln 3)W(t)\).

Answer:

\(-\ln 3\)